Question

Two beads, each with charge q and mass m, are on a horizontal, frictionless, non-conducting, circular hoop of radius R. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [ $\epsilon_0$is the permittivity of free space.]

• A. $\frac{q^2}{4\pi\epsilon_0R^3m}$
• B. $\frac{q^2}{32\pi\epsilon_0R^3m}$
• C. $\frac{q^2}{8\pi\epsilon_0R^3m}$
• D. $\frac{q^2}{16\pi\epsilon_0R^3m}$

Answer 1

The Correct Option is B

Oscillating Bead and Coulomb Force 

1. Coulomb Force Acting on the Bead

The force $F$ between the two charges is given by Coulomb’s law:

$$F = \frac{k q^2}{(2R)^2} = \frac{q^2}{4\pi \varepsilon_0 (2R)^2}.$$

2. Restoring Force Along the Tangent

The restoring force $F_R$ along the tangent at the bead’s position is:

$$F_R = -F \sin\theta,$$

where $\theta$ is the angular displacement from equilibrium.

Using the small-angle approximation, $\sin\theta \approx \theta$, and $\theta = \frac{x}{R}$, we get:

$$F_R = -\frac{q^2}{4\pi \varepsilon_0 (2R)^2} \cdot \frac{x}{R}.$$

3. Acceleration of the Bead

The acceleration $a$ of the bead is given by:

$$a = \frac{F_R}{m} = -\frac{q^2}{4\pi \varepsilon_0 (2R)^2 m} \cdot \frac{x}{R}.$$

Simplifying:

$$a = -\frac{q^2}{32\pi \varepsilon_0 R^3 m} x.$$

4. Angular Frequency for Small Oscillations

The equation of motion for simple harmonic motion is:

$$a = -\omega^2 x.$$

Comparing with our equation:

$$\omega^2 = \frac{q^2}{32\pi \varepsilon_0 R^3 m}.$$

Final Answer:

$$\frac{q^2}{32\pi \varepsilon_0 R^3 m}$$